Answer:
The force between the ball and the bat = 2233 N
Explanation:
Force: This can be defined as the product of force and its acceleration.
The S.I unit of Force is Newton (N)
F = ma .............................. Equation 1
Where F = Force , m = mass of the ball, a = acceleration of the ball.
where
a = (v-u)/t..................... Equation 2
Where v = final velocity of the ball, u = initial velocity of the ball, t = time of contact between the ball an the bat.
Given: v = 46 m/s u = - 31 m/s (it hit an horizontal line drive back at the pitcher), t = 5×10⁻³ s
Substituting these values into equation 2,
a = [46-(31)]/(5×10⁻³ )
a = 77×10³/5
a = 15.4×10³ m/s².
Also given m = 0.145 kg
Substituting into equation 1,
F = 0.145(15.4×10³ )
F = 2233 N
Thus the force between the ball and the bat = 2233 N
Force can also be defined as the product of mass and its acceleration. The force between the ball and the bat is 2233 N.
The formula of the force,
F = ma .............................. 1
Where
F = Force ,
m = mass of the ball = 0.145 kg
a = acceleration of the ball.
Formula of acceleration,
[tex]\bold {a = \dfrac {v-u}t}[/tex].................... 2
Where,
v = final velocity of the ball = 46 m/s
u = initial velocity of the ball = - 31 m/s
t = time of contact between the ball an the bat = 5×10⁻³ s
Put the values into equation 2,
[tex]\bold {a = {46-(31)}{(5x 10^{-3}} )}\\\\\bold {a = \dfrac {77x10^3}{5}}\\\\\bold {a = 15.4x10^3\ m/s^2.}[/tex]
Put the value of a into equation 1,
F = 0.145(15.4×10³ )
F = 2233 N
Therefore, the force between the ball and the bat is 2233 N.
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